Question

Two like and equal charges are at a distance of r = 5 cm, and exert a force of F
= 9×10-3 N on each other. Find: The magnitude of each charge?

Answer 1

Given:

Two like and equal charges are at a distance of r = 5 cm, and exert a force of F = 9×10-3 N on each other.

To find:

Magnitude of charges?

Calculation:

• Let the charges be denoted as 'q'.
• And Coulomb's Constant be 'k'.

$\rm \: force = \dfrac{k(q1)(q2)}{ {r}^{2} }$

$\rm \implies\: force = \dfrac{k {q}^{2} }{ {r}^{2} }$

$\rm \implies\: 9 \times {10}^{3} = \dfrac{(9 \times {10}^{9} ){q}^{2} }{ {( \frac{5}{100}) }^{2} }$

$\rm \implies\: {10}^{ - 6} = \dfrac{{q}^{2} }{ {( \frac{5}{100}) }^{2} }$

$\rm \implies\: {10}^{ - 6} = \dfrac{{q}^{2} \times {10}^{4} }{25 }$

$\rm \implies\: {q}^{2} = 25 \times {10}^{ - 10}$

$\rm \implies\: q = \sqrt{ 25 \times {10}^{ - 10} }$

$\rm \implies\: q =5 \times {10}^{5} \: C$

So, charge is 5 × 10⁵ Coulomb.

Answer 2

Answer:

$Given:</p><p></p><p>Two like and equal charges are at a distance of r = 5 cm, and exert a force of F = 9×10-3 N on each other.</p><p></p><p>To find:</p><p></p><p>Magnitude of charges?</p><p></p><p>Calculation:</p><p></p><p>Let the charges be denoted as 'q'.</p><p></p><p>And Coulomb's Constant be 'k'.</p><p></p><p>\rm \: force = \dfrac{k(q1)(q2)}{ {r}^{2} }force=r2k(q1)(q2)</p><p></p><p>\rm \implies\: force = \dfrac{k {q}^{2} }{ {r}^{2} }⟹force=r2kq2</p><p></p><p>\rm \implies\: 9 \times {10}^{3} = \dfrac{(9 \times {10}^{9} ){q}^{2} }{ {( \frac{5}{100}) }^{2} }⟹9×103=(1005)2(9×109)q2</p><p></p><p>\rm \implies\: {10}^{ - 6} = \dfrac{{q}^{2} }{ {( \frac{5}{100}) }^{2} }⟹10−6=(1005)2q2</p><p></p><p>\rm \implies\: {10}^{ - 6} = \dfrac{{q}^{2} \times {10}^{4} }{25 }⟹10−6=25q2×104</p><p></p><p>\rm \implies\: {q}^{2} = 25 \times {10}^{ - 10}⟹q2=25×10−10</p><p></p><p>\rm \implies\: q = \sqrt{ 25 \times {10}^{ - 10} }⟹q=25×10−10</p><p></p><p>\rm \implies\: q =5 \times {10}^{5} \: C⟹q=5×105C</p><p></p><p>So, charge is 5 × 10⁵ Coulomb.</p><p></p><p>$